Xenon reacts with fluorine at 873 k and 7 bar to form $Xe{F}_4$. Ratio of Xenon and fluorine required here is

At a temperature of about 873 K and pressure at 7 bar Xenon undergo reaction with 2 moles of fluorine molecules to form ${\mathrm{XeF}}_4$ molecule. The ratio of $\mathrm{Xe}:\mathrm{F}$ required for the formation of ${\mathrm{XeF}}_4$ molecule is  $1:5$. The reaction is as follows 

                $\mathrm{Xe}+2{\mathrm{F}}_2\to {\mathrm{XeF}}_4$